Exercise 12, § 6

a) Let $g$ be a finite numerical function defined on a compact interval $I = [a, b]$ of $\mathbf{R}$. One calls right-expansion set (resp. left-expansion set) of $g$ in $I$ the set of $x \in I$ such that there exists $y \in I$ for which $x < y$ (resp. $x > y$) and $g(x) < g(y)$. Show that if $g$ is continuous, then the right- (resp. left-) expansion set of $g$ in $I$ is open in $I$ and that, if $]\alpha, \beta[$ is a connected component$^1$ of this set, then $g(\alpha) = g(\beta)$ and $g(x) \leq g(\alpha)$ for $\alpha < x < \beta$.

b) Let $r_1, r_2$ be two real numbers such that $0 \leq r_1 < r_2$. Let $g$ be an increasing continuous function on $I$; let $E'$ be the left-expansion set of $g(x) - r_1 x$ in $I$, $E''$ the union of the right-expansion sets of $g(x) - r_2 x$ in each of the intervals that are the closures of the connected components of $E'$. If $\mu$ is Lebesgue measure on $I$, show, using a), that $\mu(E'') \leq \frac{r_1}{r_2} \mu(E')$.

c) One calls the upper and lower right derivates at a point $x \in I$, of a finite numerical function $f$ defined on $I$, the respective numbers

Similarly, one calls the upper and lower left derivates of $f$ at the point $x$, the respective numbers

Show that if $f$ is continuous and increasing, then the set of $x \in I$ where $D_r^+ f(x) = +\infty$ and the set of $x \in I$ where $D_l^- f(x) < D_r^+ f(x)$ have measure zero for $\mu$ (for every rational number $r > 0$ (resp. for every pair $(r_1, r_2)$ of rational numbers such that $0 \leq r_1 < r_2$), consider the set of $x \in I$ such that $D_r^+ f(x) > r$ (resp. $D_l^- f(x) < r_1$ and $D_r^+ f(x) > r_2$ simultaneously), and apply b)). From this, deduce that for almost every $x \in I$, $f$ admits a finite derivative (Lebesgue’s theorem) (apply the preceding result also to $-f(-x)$).

$^1$ Composante connexe, also translated as “component” (GT, I, §11, No. 5).